Hi,
Thank you for such a detailed description and for attaching your data file.
The way you have carried out the analysis is absolutely fine for stress calculations but crack width calculations will not be correct. This is because ASDB is trying to carry out the crack with calculations based upon the additional strains of the second stage analysis which you can see in the strain diagram and the iteration results are zero

The stress diagram is ok as the stresses are derived from the second stage strains + the applied internal strains. You can see these "summed" strains in the printed results output:

which should be almost the same as the applied internal strains but is in fact the total strain as the second stage strain is zero.
So, crack width calcs should not be performed on the second stage directly in ASBD and it is not correct to calculate the crack widths from adding the calculated crack widths from individual stage analyses without initial strains as the stress strain relationship is non-linear. It is quite possible that stage 1 loading on its own and stage two loading on its own will both result in an uncracked section - so the total will be zero - but the total applied loading will produce a cracked section.
However, it is possible use the total max/min strains in the stage 1 section, from the second stage analysis, to establish an equivalent moment and force that can be applied to the first stage section to define an equivalent second load case. This then provides the correct equivalent stress and strain in this part of the section which can be used directly in the crack width calcs.
The tricky bit is calculating the equivalent moments and axial force from the stage 2 strains but can easily be done from the analysis iteration control form but just setting the top an bottom strains of the section. It is also necessary to use two separate identical concrete materials, one for stage 1 and the other for stage 2 so that the max and min strains for stage 1 section can be extracted from the results.
In your section I have added a second load case to represent an additional 50kNm My moment to represent SDL so the total applied moment is 136kNm. The analysis of this loadcase then provides the total strains in section 1 from the results as:

If the first section analysis form is opened and the iteration form control is entered the max and min strains can be entered and the equivalent moment and force extracted.

This moment and force is then entered into an additional load case from which the crack widths can be calculated.


I appreciate that this is not straightforward but this is the way I approach it. I hope that this has helped and answered your query. If it has please mark my reply as a solution so that others may benefit. Thanks
Kind Regards
Dave Geeves